SHUFFLIN - Editorial

PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3

Author: Soumyadeep Pal
Tester: Istvan Nagy
Editorialist: Aman Dwivedi

DIFFICULTY:

Simple

PREREQUISITES:

Maths, Observation

PROBLEM:

Chef is given an array A consisting of N positive integers. Chef shuffles the array A and creates a new array B of length N, where B_i = (A_i + i) \bmod 2, for each i\;(1 \leq i \leq N).

Find the maximum possible sum of integers of the array B, if Chef shuffles the array A optimally.

EXPLANATION:

As we are taking mod by 2, then the first basic observation that we can draw from this operation is that every element of the array B is either 0 or 1.

Now the element B_i will be 1 when (A_i + i) is odd, otherwise B_i will be 0. Now recall the basic facts when is the sum of two numbers odd or even.

even +even = even \\ odd+odd=even \\ odd+even=even

Since our goal is to maximize the sum of integers of the array B we will try to shuffle array A in such a way that the number of indices where A_i and i are of opposite parity is maximized.

Hence if the element of the array A is odd we will put this element into an even index if possible. Similarly if the element of the array A is even, we will put this element into an odd index possibly.

Let E denotes the count of the even elements present in an array A, while O denotes the count of odd elements present in an array A.

Now our goal is to place the even elements into odd indices. The number of odd indices in an array of length N is:

\lceil \frac{N}{2} \rceil

Hence the number of odd indices which will contain the even elements are:

c_1 = min(E,\lceil \frac{N}{2} \rceil)

Similarly, the number of even indices present in the array of length N is:

\lfloor \frac{N}{2} \rfloor

Hence the number of odd indices which will contain the even elements are:

c_2 = min(O,\lfloor \frac{N}{2} \rfloor)

So, c_1+c_2 is the count of the number of indices where A_i and i are of opposite parity. Hence the maximum sum of array B is c_1+c_2.

TIME COMPLEXITY:

O(N) per test case.

SOLUTIONS:

Author
#include<bits/stdc++.h>
using namespace std;

void solve(int tc) {
	int n; cin >> n;
	int odd = 0, even = 0;
	for (int i = 0; i < n; i++) {
		int x; cin >> x;
		odd += (x % 2 == 1);
		even += (x % 2 == 0);
	}
	int ans = min((n + 1) / 2, even) + min(n / 2, odd);
	cout << ans << '\n';
}

signed main() {
	ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t; cin >> t;
	for (int i = 1; i <= t; i++) solve(i);
	return 0;
}

Tester
#include <iostream>
#include <cassert>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <random>

#ifdef HOME
#include <windows.h>
#endif

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

template<class T> bool umin(T& a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool umax(T& a, T b) { return a < b ? (a = b, true) : false; }

using namespace std;

long long readInt(long long l, long long r, char endd) {
	long long x = 0;
	int cnt = 0;
	int fi = -1;
	bool is_neg = false;
	while (true) {
		char g = getchar();
		if (g == '-') {
			assert(fi == -1);
			is_neg = true;
			continue;
		}
		if ('0' <= g && g <= '9') {
			x *= 10;
			x += g - '0';
			if (cnt == 0) {
				fi = g - '0';
			}
			cnt++;
			assert(fi != 0 || cnt == 1);
			assert(fi != 0 || is_neg == false);

			assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
		}
		else if (g == endd) {
			assert(cnt > 0);
			if (is_neg) {
				x = -x;
			}
			assert(l <= x && x <= r);
			return x;
		}
		else {
			assert(false);
		}
	}
}

string readString(int l, int r, char endd) {
	string ret = "";
	int cnt = 0;
	while (true) {
		char g = getchar();
		assert(g != -1);
		if (g == endd) {
			break;
		}
		cnt++;
		ret += g;
	}
	assert(l <= cnt && cnt <= r);
	return ret;
}
long long readIntSp(long long l, long long r) {
	return readInt(l, r, ' ');
}
long long readIntLn(long long l, long long r) {
	return readInt(l, r, '\n');
}
string readStringLn(int l, int r) {
	return readString(l, r, '\n');
}
string readStringSp(int l, int r) {
	return readString(l, r, ' ');
}

int main(int argc, char** argv)
{
#ifdef HOME
	if (IsDebuggerPresent())
	{
		freopen("../in.txt", "rb", stdin);
		freopen("../out.txt", "wb", stdout);
	}
#endif
	int T = readIntLn(1, 10'000);
	int sumN = 0;
	forn(tc, T)
	{
		int N = readIntLn(1, 100'000);
		sumN += N;
		vector<int> A(N);
		int odd = 0, even = 0;
		forn(i, N)
		{
			if (i + 1 == N)
				A[i] = readIntLn(0, 1'000'000'000);
			else
				A[i] = readIntSp(0, 1'000'000'000);
			if (A[i] & 1)
				++odd;
			else
				++even;
		}
		int res = min<int>((N - N / 2), even) + min<int>(N / 2, odd);
		printf("%d\n", res);
	}
	assert(sumN < 300'000);
	assert(getchar() == -1);
	return 0;
}

Editorialist
#include<bits/stdc++.h>
using namespace std;

#define int long long

void solve()
{
	int n;
	cin>>n;

	int even = 0,even_ind=0;
	int odd = 0,odd_ind=0;

	int a[n];

	for(int i=0;i<n;i++)
	{
		cin>>a[i];

		if(a[i]%2)
			odd++;
		else
			even++;

		if(i%2)
			even_ind++;
		else
			odd_ind++;
	}

	int ans = min(odd_ind,even)+min(even_ind,odd);

	cout<<ans<<endl;
}

int32_t main()
{
	// freopen("input.txt","r",stdin);
	// freopen("output.txt","w",stdout);

	int t;
	cin>>t;

	while(t--)
		solve();

return 0;
}

as far as my knowledge goes even+odd=odd . editorial state otherwise might be a typo

I have another way to solved,I solved it ,but codechef return wrong answer.But i did not find any case where my method goes wrong.Can anyone test my method.